Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm working through my book, on the section about compound angle formulae. I've been made aware of the identity $\sin(A + B) \equiv \sin A\cos B + \cos A\sin B$. Next task was to replace B with -B to show $\sin(A - B) \equiv \sin A\cos B - \cos A \sin B$ which was fairly easy. I'm struggling with the following though:

"In the identity $\sin(A - B) \equiv \sin A\cos B - \cos A\sin B$, replace A by $(\frac{1}{2}\pi - A)$ to show that $\cos(A + B) \equiv \cos A\cos B - \sin A\sin B$."

I've got $\sin((\frac{\pi}{2} - A) - B) \equiv \cos A\cos B - \sin A\sin B$ by replacing $\sin(\frac{\pi}{2} - A)$ with $\cos A$ and $\cos(\frac{\pi}{2} - A)$ with $\sin A$ on the RHS of the identity. It's just the LHS I'm stuck with and don't know how to manipulate to make it $\cos(A + B)$.

P.S. I know I'm asking assistance on extremely trivial stuff, but I've been staring at this for a while and don't have a tutor so hope someone will help!

share|improve this question
2  
Hint $\sin((\frac{\pi}{2} - A) - B) =\sin((\frac{\pi}{2} - (A + B)) $ and what is $\sin (\frac{\pi}{2}-$angle$)$? –  N. S. May 24 '11 at 21:56
    
please do accept an answer. –  user9413 May 25 '11 at 6:18
    
@Chandru: Answer accepted now. (N.B. I am in the UK and posted this just before going to bed). –  PeteUK May 25 '11 at 9:55
    
Hey no problem.I just reminded you :) –  user9413 May 25 '11 at 9:55
    
@user9176: This is not the first time I've been stuck on a trigonometry problem when the answer is reachable via some trivial algebraic manipulation. Thanks for the comment. –  PeteUK May 25 '11 at 9:57

4 Answers 4

up vote 5 down vote accepted

$$\sin\left(\left(\frac{\pi}{2} - A\right) - B\right) =\sin\left(\frac{\pi}{2} - (A+B) \right)= \cos (A+B)$$

share|improve this answer
    
Very clear and concise answer. Thanks! –  PeteUK May 25 '11 at 9:54

Note that you can also establish:

$$\sin\left(\left(\frac{\pi}{2} - A\right) - B\right) =\sin\left(\frac{\pi}{2} - (A + B)\right) = \cos(A+B)$$ by using the second identity you figured out above, $\sin(A - B) \equiv \sin A\cos B - \cos A\sin B$, giving you:

$$\sin\left(\left(\frac{\pi}{2} - A\right) - B\right) = \sin\left(\frac{\pi}{2} - (A+B)\right)$$ $$ = \sin\left(\frac{\pi}{2}\right)\cos(A+B) - \cos\left(\frac{\pi}{2}\right)\sin(A+B)$$ $$= (1)\cos(A+B) - (0)\sin(A+B)$$ $$ = \cos(A+B)$$

share|improve this answer
    
thanks. I enjoyed the extra bit going through the second identity too. –  PeteUK May 25 '11 at 9:43

This is the same as Henry's answer, only presented differently.
$\left(\displaystyle \frac{\pi}{2}-A\right) - B = \displaystyle \frac{\pi}{2} - (A+B)$. Now you can use the fact that $\sin \left(\displaystyle \frac{\pi}{2} - C\right) = \cos(C)$...

share|improve this answer
    
Appreciate your answer. –  PeteUK May 25 '11 at 9:53

Update: Since $\sin (\frac{\pi }{2}-x)\equiv \cos x$, you have

$$\sin \left(\frac{\pi }{2}-A-B\right)\equiv \sin \left(\frac{\pi }{2}-(A+B)\right)\equiv \cos (A+B).$$


Replacing $A$ by $\frac{\pi }{2}-A$ in

$$\sin (A-B)\equiv \sin A\cos B-\cos A\sin B,$$

gives

$$\sin \left(\frac{\pi }{2}-A-B\right)\equiv \sin \left( \frac{\pi }{2}-A\right) \cos B-\cos \left( \frac{\pi }{2}-A\right) \sin B.$$

Since $\sin (\frac{\pi }{2}-x)\equiv \cos x$ and $\cos (\frac{\pi }{2}-x)\equiv \sin x$, you have

$$\sin \left(\frac{\pi }{2}-A-B\right)\equiv \sin \left(\frac{\pi }{2}-(A+B)\right)\equiv \cos (A+B),$$

$$\sin \left( \frac{\pi }{2}-A\right) \equiv \cos A,$$

and

$$\cos \left( \frac{\pi }{2}-A\right) \equiv \sin A.$$

Hence

$$\cos (A+B)\equiv \cos A\cos B-\sin A\sin B.$$

share|improve this answer
    
érico: thank you very much for helping me out again. –  PeteUK May 25 '11 at 9:52

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.